 Hello, and welcome to this screencast. In this screencast, we will determine whether or not an infinite series converges using the integral test. So here is the statement of the integral test for the convergence of a series. Let f be a real valued function that is positive and decreasing for all x larger than some value c. And suppose a sub k, the k-th term of our series, is equal to f of k for each positive integer k. Then the series from n equals 1 to infinity of a sub n converges if and only if the improper integral from c to infinity of f of x dx converges. So this means we need to create an appropriate improper integral, and then we determine if that integral converges or diverges. The outcome for our integral will be the same as the outcome for our original series, and that's how we'll answer the question. So let's start by writing our general term a sub n equals 1 over n to the fourth. To create a function f of x, we will just replace n with x, so we will use f of x equals 1 over x to the fourth. So now let's check to make sure that this function f of x satisfies the conditions of the integral test. So it needs to be a real valued function, and it certainly is. All of the outputs are going to be real numbers. And it also needs to be positive and decreasing for all x larger than some value c. So 1 over x to the fourth is always going to be positive for any input except for 0. And so that will work. And if we let c be equal to 1, then as x increases, the denominator will get larger. So 1 over x to the fourth is also going to decrease. So now we want to check that the kth term of the series is equal to f of k, and so that's going to be true. That's exactly how we created that function f of x. And so this satisfies the hypothesis of the integral test. And so now in order to use this, we need to see if the improper integral converges or diverges. That's our next step. So our improper integral for this example is the integral from 1 to infinity of 1 over x to the fourth dx. Our first step in evaluating this integral is to replace the upper limit with a variable that represents a finite value b. And then we're going to evaluate that limit as b goes to infinity. And in this step, we're also going to rewrite the integrand as x to the negative fourth dx to make it easier to work with. So to find an antiderivative, we can just use the power rule. And so we'll have a limit as b approaches infinity of x to the negative three divided by that new exponent negative three. And we're going to evaluate that at b and at 1. So when we evaluate this at x equals b, we get 1 over negative 3 b cubed. I moved that back down to the denominator so we'd have a positive exponent. And then when we evaluate at x equals 1, we're going to subtract 1 over negative 3. And so to evaluate this limit now, let's consider this term here, the 1 over negative 3 b cubed. As b increases, this denominator is also going to increase. It's going to become very, very large. And that means the whole fraction, 1 over negative 3 b cubed, is going to become very, very small. And so this part of it will approach 0 as b approaches infinity. So now we're left with 0 minus negative 1, 0 minus 1 over negative 3. And so that simplifies to just 1 third. And so that's the value of our limit as b approaches infinity. So since the limit exists, that means that the improper integral converges. So we figured out from our integral test whether or not the integral converges. And we know that the outcome for the series is going to be the same outcome for our improper integral using the integral test. And so that means that since our improper integral converges, the series from n equals 1 to infinity of 1 over n to the 4th also converges. And so that's our final result using the integral test. Thanks for watching.